= Solution
Let $R$ be an <orthogonal matrix> describing a Cartesian change of coordinates, with transformed <vector> components $v_i'=R_{ia}v_a$ and $w_j'=R_{jb}w_b$. Their <outer product> transforms as
$$
T_{ij}'=v_i'w_j'=R_{ia}R_{jb}T_{ab},
$$
which is the transformation law of a second-order Cartesian <tensor>. Here “rank two” counts tensor indices, not the <rank of a matrix>; a nonzero <outer product> has matrix rank one.
For an <isotropic second-rank tensor>, invariance means $RTR^T=T$ for every proper <rotation matrix>. The half-turn matrices $\operatorname{diag}(1,-1,-1)$, $\operatorname{diag}(-1,1,-1)$ and $\operatorname{diag}(-1,-1,1)$ force all off-diagonal entries to vanish. Quarter-turns around coordinate axes then force the three diagonal entries to be equal. Conversely $RIR^T=I$, so \b[the most general second-rank <isotropic tensor> is]
$$
\boxed{T_{ij}=\lambda\delta_{ij},\qquad\lambda\in\mathbb R.}
$$
Since $vw^T$ has matrix rank at most one, it cannot equal a nonzero scalar multiple of the three-dimensional identity. Thus its isotropy forces $vw^T=0$. If $v\ne0$, a nonzero component $v_i$ makes the entire $i$th row vanish only when $w=0$; the converse is immediate. \b[The required choices are exactly]
$$
\boxed{v=0\quad\text{or}\quad w=0.}
$$
For the <Levi-Civita symbol>, multilinearity and antisymmetry of the <determinant> give
$$
R_{ia}R_{jb}R_{kc}\epsilon_{abc}=(\det R)\epsilon_{ijk}=\epsilon_{ijk}
$$
for every $R\in SO(3)$. This proves the <rotation invariance of the Levi-Civita symbol>, hence its third-rank isotropy. The convention is invariance under proper rotations: under an orthogonal reflection $\det R=-1$, the sign reverses. Thus it is a <pseudotensor> if transformations of both orientations are included.
Back to article page
